91Ó°ÊÓ

The given equation is$f\left(x\right)=\left(\begin{array}{c}x,\\ 2-x,\\ 0\end{array}\begin{array}{ccccc}0& ≤& x& ≤& 1\\ 1& ≤& x& ≤& 2\\ & x& â‰\yen & 2& \end{array}\right)$

An infinite sum of sines and cosines is used to represent the expansion of a periodic function f(x) into a Fourier series. The orthogonality relationships between the sine and cosine functions are used in the Fourier series.

The Fourier cosine transform is equal to:

$$\begin{gathered} g_c\mathrm{Î\pm }=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{1}x\cos \left(\mathrm{Î\pm }x\right)dx+\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_1^2\left(2-x\right)\cos \left(\mathrm{Î\pm }x\right)dx \\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\left[\left(\frac{x}{\mathrm{Î\pm }}\sin \left(\mathrm{Î\pm }x\right)+\frac{1}{{\mathrm{Î\pm }}^2}\cos \left(\mathrm{Î\pm }x\right)\right){\left|+2\frac{\sin \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }}\right|}_1^2-{\left(\frac{x}{\mathrm{Î\pm }}\sin \left(\mathrm{Î\pm }x\right)+\frac{1}{{\mathrm{Î\pm }}^2}\cos \left(\mathrm{Î\pm }x\right)\right)}_1^2\right] \\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{1}{{\mathrm{Î\pm }}^2}\left(\cos \mathrm{Î\pm }-1-\cos \left(2\mathrm{Î\pm }\right)+\cos \mathrm{Î\pm }\right) \\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{1}{{\mathrm{Î\pm }}^2}\left(2\cos \mathrm{Î\pm }-\cos \left(2\mathrm{Î\pm }\right)-1\right) \end{gathered}$$

Further solving

$$\begin{gathered} g_c\mathrm{Î\pm }=\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{1}{{\mathrm{Î\pm }}^2}\left(2\cos \mathrm{Î\pm }-2{\cos }^2\mathrm{Î\pm }\right) \\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{2}{{\mathrm{Î\pm }}^2}\cos \mathrm{Î\pm }\left(1-\cos \mathrm{Î\pm }\right) \\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{4}{{\mathrm{Î\pm }}^2}\cos \mathrm{Î\pm }{\sin }^2\frac{\mathrm{Î\pm }}{2} \end{gathered}$$

Thus, the function is equal to

$f\left(x\right)=\frac{8}{\mathrm{Ï€}}{∫}_0^{∞}\frac{\cos \mathrm{Î\pm }{\sin }^2\left(\mathrm{Î\pm }/2\right)}{{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }$

Now set x=1 , and f(1)=1 use

$$\begin{gathered} 1=\frac{8}{\mathrm{Ï€}}{∫}_0^{∞}\frac{{\cos }^2\mathrm{Î\pm }{\sin }^2\left(\mathrm{Î\pm }/2\right)}{{\mathrm{Î\pm }}^2}d\mathrm{Î\pm } \\ {∫}_0^{∞}\frac{{\cos }^2\mathrm{Î\pm }{\sin }^2\left(\mathrm{Î\pm }/2\right)}{{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }=\frac{\mathrm{Ï€}}{8} \end{gathered}$$

Therefore,the cosine transform is$f\left(x\right)=\frac{2}{\mathrm{Ï€}}{∫}_0^{∞}\frac{\cos \mathrm{Î\pm }{\sin }^2\left(\mathrm{Î\pm }/2\right)}{{\mathrm{Î\pm }}^2}\cos \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$

and the value of the integral is

${∫}_0^{∞}\frac{{\cos }^2\mathrm{Î\pm }{\sin }^2\left(\mathrm{Î\pm }/2\right)}{{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }=\frac{\mathrm{Ï€}}{8}$